Another method for inverse Z-transforms is partial fraction decomposition. Lets start by example.

\begin{align*}
\hat{H}(z) &= \frac{z - \frac{3}{2}}{ \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} }} \quad h \mbox{ is causal } \\
\end{align*}

Since the system is causal and the largest pole is $\frac{3}{4}$, we know that the $RoC = \{ z \st \abs{z} \gt \frac{3}{4} \}$, and this includes the unit circle which implies BIBO stability. In this case, $\hat{H}(z)$ is strictly proper and the poles are unique. This is the most straight forward case of partial fraction decomposition.

\begin{align*}
\hat{H}(z) &= \frac{z - \frac{3}{2}}{ \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} }}  \\
\hat{H}(z) &= \frac{A}{ \lr{ z + \frac{1}{2}}} + \frac{B}{ \lr{ z - \frac{3}{4} }}  \\
\hat{H}(z) &= \frac{Az^{-1}}{ \lr{ 1 + \frac{1}{2}z^{-1}}} + \frac{Bz^{-1}}{ \lr{ 1 - \frac{3}{4}z^{-1} }}  \\
h(n) &=  A\lr{-\frac{1}{2}}^{n-1} u(n-1) + B\lr{\frac{3}{4}}^{n-1} u(n-1)
\end{align*}

Note that if the order of the numerator is greater than the order of the denominator, then the $RoC$ doesn't include infinity which implies that the system cannot be causal. To solve for $A$ and $B$, we have to continue with the partial fraction decomposition to set up a system of equations:

\begin{align*}
\hat{H}(z) &= \frac{A}{ \lr{ z + \frac{1}{2}}} + \frac{B}{ \lr{ z - \frac{3}{4} }}  \\
\hat{H}(z) &= \frac{ A\lr{z - \frac{3}{4}} + B\lr{ z + \frac{1}{2} } } {  \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} }} \\ 
\frac{z - \frac{3}{2}}{ \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} }} &= \frac{ A\lr{z - \frac{3}{4}} + B\lr{ z + \frac{1}{2} } } {  \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} }} \\ 
z - \frac{3}{2} &= A\lr{z - \frac{3}{4}} + B\lr{ z + \frac{1}{2} } \\
z - \frac{3}{2} &= z\lr{A + B} + -\frac{3}{4}A + \frac{1}{2}B  \\
\end{align*}

Equating the coefficients gives us the system of equations:
\begin{align*}
A + B &= 1 \\
-\frac{3}{4}A + \frac{1}{2}B &= -\frac{3}{4} \\
\end{align*}

Solving the system gives $A = \frac{8}{5}$, and $B = -\frac{3}{5}$. Putting this all together gives us $\hat{H}(z)$:

$$
\hat{H}(z) = \frac{\frac{8}{5}}{z + \frac{1}{2}} - \frac{ \frac{3}{5} }{ z - \frac{3}{4} } \ztp \frac{8}{5}\lr{-\frac{1}{2}}^{n-1}u(n-1) - \frac{3}{5}\lr{\frac{3}{4}}^{n-1} u(n-1) 
$$

A partial fraction shortcut is to use the Heaviside method.

\begin{align*}
\hat{H}(z) &=  \frac{A}{ \lr{ z + \frac{1}{2}}} + \frac{B}{ \lr{ z - \frac{3}{4} }}  \\
\lr{z + \frac{1}{2}} \hat{H}(z) &=  A + \frac{B\lr{z + \frac{1}{2}}}{ \lr{ z - \frac{3}{4} }}  \\
\left. \lr{z + \frac{1}{2}} \hat{H}(z) \right|_{z=-\frac{1}{2}} &=  A \\
\end{align*}

To solve for $B$ just to the corresponding multiplication and evaluation. Isolate the solution of each parameter instead of solving systems.

\begin{example}
Find the inverse transform of the following system.
\begin{align*}
\hat{X}(z) &= \frac{ 1 + 2z^{-1} + z^{-2}}{\lr{ 1 - \frac{1}{2}z^{-1}} \lr{ 1 - z^{-1}}} \quad \abs{z} \gt 1 \\
\end{align*}

Note that the order of the numerator is equatl to the order of the denominator. This causes some issues when using partial fraction decomposition. One method is to add an addition coefficient into the system:

\begin{align*}
\hat{X}(z) &= A + \frac{B}{z - \frac{1}{2}} + \frac{C}{z-1} \\
x(n) &= A\delta(n) + B\lr{\frac{1}{2}}^{n-1} u(n-1) + C u(n-1) \\
\end{align*}

We can use the initial value theorem to determine $A$ if $x$ is causal. $x(0) = \lim_{z\to\infty} \hat{X}(z) = 1$. Then use the Heaviside method to solve for $B$ and $C$. Another method to find these coefficients in a more algorithmic manner is long division.

\begin{center}
\polyset{vars=z}
\polylongdiv{z^2 + 2z + 1}{z^2 -\frac{3}{2}z + \frac{1}{2}}
\end{center}

\begin{align*}
\hat{X}(z) &= \frac{ 1 + 2z^{-1} + z^{-2}}{\lr{ 1 - \frac{1}{2}z^{-1}} \lr{ 1 - z^{-1}}} \\
\hat{X}(z) &= \frac{ z^2 + 2z + 1}{\lr{ z - \frac{1}{2}} \lr{ z - 1}} \\
\hat{X}(z) &= 1 + \frac{7z + 1}{2\lr{z^2 - \frac{3}{2}z + \frac{1}{2}}} \quad \mbox{ after division } \\
\hat{X}(z) &= 1 + \frac{7z + 1}{2\lr{z-\frac{1}{2}}\lr{z-1}}  \\
\end{align*}

Now we have the first coefficient $A = 1$ and a proper rational function in $z$. Now we have a system which we already know how to solve for---using partial fraction decomposition or the shortcut with the Heaviside method.

\end{example}

\begin{example}

Consider the following problem:

\begin{align*}
\hat{G}(z) &= \frac{ z^3 + 2z^2 + z }{ z^2 -\frac{3}{2}z + \frac{1}{2} } \\
\end{align*}

The system is not causal because the order of the numerator is higher than the denominator. This is because this arrangement implies that there are poles at $\infty$. Nevertheless, we can still solve this problem:

\begin{align*}
\hat{G}(z) &= \frac{ z^3 + 2z^2 + z }{ z^2 -\frac{3}{2}z + \frac{1}{2} } \\
\hat{G}(z) &= Az + B + \frac{C}{z - \frac{1}{2}} + \frac{D}{z - 1} \\
g(n) &= A\delta(n-1) + B\delta(n) + C\lr{\frac{1}{2}}^{n-1} + D u(n-1)\\
\end{align*}

How do we find the coefficients $A$ and $B$? Long division will ameliorate the problem. 

\begin{center}
\polyset{vars=z}
\polylongdiv{z^3 + 2z^2 + z}{z^2 -\frac{3}{2}z + \frac{1}{2}}
\end{center}

\begin{align*}
\hat{G}(z) &= \frac{ z^3 + 2z^2 + z }{ z^2 -\frac{3}{2}z + \frac{1}{2} } \\
\hat{G}(z) &=  z + \frac{7}{2} + \frac{ 23z - 7 }{ 4\lr{z - \frac{1}{2}} \lr{z-2}} \quad \mbox{after division} \\
\end{align*}

This gives us $A=1$ and $B = \frac{7}{2}$. Now we just have to solve for $C$ and $D$ using the equation:

\begin{align*}
\frac{ 23z - 7 }{ 4\lr{z - \frac{1}{2}} \lr{z-2}} &=  \frac{C}{ \lr{z - \frac{1}{2}}} + \frac{D}{ \lr{z-2}} \\
\end{align*}

Now just solve for $C$ and $D$ as usual.

\end{example}

\begin{example}

Suppose we have a problem with duplicate poles.

\begin{align*}
\hat{F}(z) &= \frac{ \frac{2}{3} z^2 -\frac{1}{3}z = \frac{1}{2} }{\lr{z - \frac{1}{2}}^2 \lr{z - 2}} \\
\end{align*}

Assume that the system is causal. How do you find the inverse Z-transform? We can break it up into fractions with constant coefficients for each power in the denominator:

\begin{align*}
\hat{F}(z) &= \frac{A}{z-\frac{1}{2}} + \frac{B}{\lr{z - \frac{1}{2}}^2} + \frac{C}{z - 2} \\
\end{align*}

We can use a combination of the Heaviside method and regular partial fraction decomposition to solve this problem:

\begin{align*}
B &= \left. \lr{z - \frac{1}{2}} \hat{F}(z) \right|_{z = \frac{1}{2}} \\
C &= \left. \lr{z - 2} \hat{F}(z) \right|_{z = 2} \\
\end{align*}

To find $A$, multiply all terms until we have common denominator on both sides and compare coefficients of the numerators:

\begin{align*}
\frac{2}{3}z^2 - \frac{1}{3}z - \frac{1}{2} &= (A + C)z^2 + \lr{\frac{5}{2}A + B - C}z + \lr{A -2B + \frac{1}{4}C}
\end{align*}

We find that the only system that matters is $A + C = \frac{2}{3}$. This is because we already know the value of $C$. Another method for getting $A$ is by using differentiation:

\begin{align*}
\hat{F}(z) &= \frac{A}{z-\frac{1}{2}} + \frac{B}{\lr{z - \frac{1}{2}}^2} + \frac{C}{\lr{z - 2}} \\
\lr{z - \frac{1}{2}}^2 \hat{F}(z) &= A\lr{z-\frac{1}{2}} + B + C\frac{\lr{z - \frac{1}{2}}^2}{\lr{z - 2}} \\
\frac{d}{dz}\lr{z - \frac{1}{2}}^2 \hat{F}(z) &= A + C\lr{\frac{2\lr{z - \frac{1}{2}}\lr{z-2} - \lr{z-\frac{1}{2}}^2 }{\lr{z - 2}^2}} \\
\left. \frac{d}{dz}\lr{z - \frac{1}{2}}^2 \hat{F}(z) \right|_{z = \frac{1}{2}}&= A \\
\end{align*}

After differentiating, we can use the Heaviside method. If we had a cubed term, we would have to differentiate again, and so on with higher powers.

\end{example}

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